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Question
I'd like to ask you one more question, since you were so knowledgeable on the last one (I'm in the process of reviewing for finals).

THE QUESTION:
If m and n are natural numbers with gcd(m,n) = 1, how can we prove that Q(ζm,ζn) = Q(ζmn) and that Q(ζm)∩Q(ζn) = Q?

(I understood the notation as Q=rationals, and Q(ζm,ζn) denotes the splitting field obtained by adjoining primitive nth and mth roots of unity, and Q(ζmn) denotes the cyclotomic field of (mn)th roots of unity).

My thoughts:
I thought it would be easiest to let ζn=e^(2πi/n) and ζm=(2πi/m), that is, the first nth and mth roots of unity respectively counterclockwise from 1. So,
(ζmn)^m=e^(2πmi/mn)=ζn
and
(ζmn)^n=e^(2πni/mn)=ζm

From this, can we say that Q(ζmn) contains Q(ζm,ζn)??

Inclusion the other direction needs to be shown as well, i.e. it must be shown that Q(ζm,ζn) contains Q(ζmn).

I know it's important that m and n are relatively prime. This tells us that if m<n, then (ζn)^m is going to be another primitive nth root of unity, or vice versa if n<m.

I think part of what I'm struggling with is what exactly Q(ζm,ζn) "looks like."

I suppose, for example, elements of Q(ζn) would be as follows:
(*)   Q(ζn)={a0+a1(ζn)+a2(ζn)^2+...+a(n-1)(ζn)^(n-1)},
with the ai elements of Q,
so elements of Q(ζm,ζn) would look like
Q(ζm,ζn)={b0+b1(ζm)+b2(ζm)^2+...+b(m-1)(ζm)^(m-1)},
where the bi are elements of Q(ζn) as described in (*).

Could we maybe say from this, if b1 just equals ζn, we will get ζmζn in Q(ζm,ζn), and ζmζn=e^(2πi(m+n)/mn)
is a primitive (mn)th root of unity, so Q(ζm,ζn) contains Q(ζmn)?



As for the second part, which is to show that Q(ζm)∩Q(ζn) = Q,
it seems like we need to think about the fact that (ζm)^j=e^(2πji/m) for j=0,1,...,m-1 and (ζn)^k=e^(2πki/n) for k=0,1,...,n-1, and the only place where we are going to get (ζm)^j=(ζn)^k is at k=j=0,
i.e. Q(ζm) and Q(ζn)don't "share" any primitive roots of unity since gcd(m,n)=1.

I apologize if my thinking is way off track; I'm struggling with this particular concept, and can't tell if I'm even remotely close to the right idea or not!

Answer
You have all the right ideas here. So let's talk through it because you're mostly there.

First of all, ζn and ζm can be any of the mth or nth primitive roots of unity, but as you say, Q(ζm) is the same regardless of which you use.

So you really want to think about is what Q(ζmn) is. It is, as you said, just Q plus an (mn)th root of unity. It is very clear that Q(ζmn) contains ζm and ζn because (ζmn)^n = ζm and similarly (ζmn)^m = ζn.

To prove that formally in two ways. You could set ζk = e^(2πi/k) and just doing the computation. You could prove it abstractly -- of course (ζmn)^n is an m-th root of unity, and if it were not primitive, then neither would be ζmn.

So that proves Q(ζm,ζn) ⊆ Q(ζmn). Now you have to prove the opposite.

If you have Q(ζm,ζn), then this should also contain all powers of ζm and ζn and any product of those. Check out the product ζm ζn. It should not be too hard to show that this is a primitive root of unity (not the usual one, not e^(2πi/mn), but that is fine).

You can also do this abstractly (and this is one way to look at the first part of this problem) you have two subgroups of the infinite group T={z : |z|=1} in the complex numbers. The intersection of T with your extension Q(ζm,ζn) is a finite group (call it G). In particular, you have two elements of G, ζm and ζn, which are of order m and n. Thus (since your whole ring, and T in particular, is commutative) you know that |G| is at least mn. You can show, without much work, that G is <ζm> x <ζn>, or more precisely, it is generated by those two roots. Thus, because m and n are coprime, it is also a cyclic group. Any generator of G will be a primitive mn root of unity within Q(ζm,ζn).


Thus Q(ζmn) ⊆ Q(ζm,ζn)




Finally, as for the intersection, you should have a theorem telling you that if you have two field extensions over Q, one of order m, the other of order n, if the largest field containing both of them is order mn, then their intersection must be Q. It might be called a "diamond lemma" or some other term. You can also prove it directly, since ζm will not show up in Q(ζn) and vice-versa. There are no elements of multiplicative order m in Q(ζn).


Sorry for the delay in my response -- I wrote out an answer for this days ago, but I don't know where it went. I must have submitted it and it didn't go through or something. What I've omitted re-typing this second time is the relationship between Q(ζm) and the cyclotomic polynomial Φm(x), which is (succinctly) that Q(ζm) = Z[x] / (Φm(x)). You could approach this problem from that point of view as well, but just looking at Q(ζm) and Q(ζn) is simpler and more hands-on.

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